How do I reciprocate a polynomial?

Algebra Level 3

$\large \displaystyle \sum_{n=1}^{2015} \dfrac{1}{T_{n}}$

If $T_{n} = \frac{1}{4} (n+2)(n+3)$ , the value of the above sum is in the form $\dfrac{A}{B}$ , where $A,B \in \mathbb{N}$ and $\gcd(A,B)=1$ . Find $A+B$ .

The answer is 7057.

1 solution

Rishabh Jain
Mar 17, 2016

$\begin{aligned}\displaystyle \sum_{n=1}^{2015} \dfrac{1}{T_{n}}&=\displaystyle \sum_{n=1}^{2015} \dfrac{4}{(n+2)(n+3)}\\&=4\displaystyle \sum_{n=1}^{2015} \left(\dfrac{1}{n+2}-\dfrac{1}{n+3}\right)\end{aligned}$ $\color{#456461}{\mathbf{(A~Telescopic ~Series)}}$ $\large =4\left(\dfrac{1}{3}-\dfrac{1}{2018}\right)$ $\large =\dfrac{4030}{3027}$ $\therefore 4030+3027=\Huge\color{#456461}{\boxed{\color{#EC7300}{\boxed{\color{#0C6AC7}{\mathbf{7057}}}}}}$

Same soln.

Aditya Narayan Sharma - 5 years, 2 months ago

Another fine solution by "The Cool"! :)

tom engelsman - 4 years, 4 months ago

Thanks... :-)

Rishabh Jain - 4 years, 4 months ago

How do I reciprocate a polynomial?

The answer is 7057.

1 solution

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